Summary: Addition is bit-wise xor. Multiplication (which you denote as $\&$), is polynomial multiplication in the field $\mathbb{F}_{2^m}$, i.e. modulo some irreducible polynomial of degree $m$ over $\mathbb{F}_2$.
We construct $\mathbb{F}_{(2^m)^2}$ in such a way that it contains an element $x$ such that $P(x)=0$. Since this is a polynomial of degree 2, and it splits in $\mathbb{F}_{(2^m)^2}$, there exist also a $y\in\mathbb{F}_{(2^m)^2}$ such that
$$X^2+X+\phi=(X-x)(X-y),$$
where $P(X)=X^2+X+\phi$ is a polynomial in $\mathbb{F}_{2^m}[X]$.
It follows that $xy=\phi$ and $x+y=-1$, hence $y=\frac{\phi}{x}$ and $y=-x-1$. Therefore
\begin{align}
(a_0+a_1x)^{-1} &= (a_0+a_1x)^{-1}(a_0+a_1y)^{-1}(a_0+a_1y) \\
&= (a_0^2+a_0a_1y+a_0a_1x+a_1^2xy)^{-1}(a_0+a_1y) \\
&= (a_0^2-a_0a_1+a_1^2\phi)^{-1}(a_0+a_1y) \\
&= (a_0(a_0-a_1)+a_1^2\phi)^{-1}(a_0-a_1x-a_1) \\
&= (a_0(a_0+a_1)+a_1^2\phi)^{-1}(a_0+a_1+a_1x). \\
\end{align}
This is how you arrive at the inverse.
Now note that $a_0,a_1,\phi\in\mathbb{F}_{2^m}$, so all operations to compute $(a_0(a_0+a_1)+a_1^2\phi)$ are in the field $\mathbb{F}_{2^m}$. In other words, addition is just bit-wise xor. Multiplication is polynomial multiplication modulo an irreducible polynomial over $\mathbb{F}_2$ of degree $m$.
Moreover, this means that the inversion $(a_0(a_0+a_1)+a_1^2\phi)^{-1}$ is also computed in the field $\mathbb{F}_{2^m}$, hence we have reduced inversion in $\mathbb{F}_{(2^m)^2}$ to an inversion in $\mathbb{F}_{2^m}$ plus some cheap stuff.
&is probably bitwise AND. $\endgroup$ – fkraiem Dec 6 '16 at 9:13&to be bitwise AND, additionally this notation should be documented somewhere in your course prior to using it here. $\endgroup$ – SEJPM♦ Dec 6 '16 at 9:47